Solving simple differential equation (t+4)dx=4(1+x2)dt

As you rearranged it in your post, we have
${\displaystyle \int \frac{dx}{1+x^2}=4\int \frac{dt}{t+4}}$
As you discovered, the LHS is arctan. To find the value of the RHS, we may say ${\displaystyle t+4=u\text{so}du=dt}$. Thus the RHS can be rewritten as
${\displaystyle 4\int \frac{du}{u}=4\log \left|u\right|+C_2}$
Because we want this integral to be in respect to t, we have ${\displaystyle u=t+4}$, so
${\displaystyle \arctan x+C_1=4\log |t+4|+C_2}$
To confirm that the integral we solved is indeed correct, you can simply differentiate ${\displaystyle 4\log |t+4|+C_2}$
Thus ${\displaystyle x=\tan (4\log |t+4|+C_2-C_1)}$
and ${\displaystyle t=\exp \left(\frac{1}{4}(\arctan x+C_1-C_2)\right)-4}$

Here are some proceeding steps:
- Seperating the variables you get
${\displaystyle \frac{dx}{1+x^2}=4\cdot \frac{dt}{t+4}}$
- Integrating both sides you have:
${\displaystyle {\tan }^{-1}\left(x\right)+C_1=4\cdot \log |t+4|+C_2}$.